Codes in the Damerau Distance for DNA Storage

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by  Ryan Gabrys

Codes in the Damerau Distance for Deletion and
Adjacent Transposition Correction
Ryan Gabrys∗, Eitan Yaakobi ‡, and Olgica Milenkovic∗
∗ECE Department, University of Illinois, Urbana-Champaign ‡Technion University
Abstract—Motivated by applications in DNA-based storage, we moreprecisely,byexposureto standardor increasedlevelradi-
introduce the new problem of code design in the Damerau met- ation, humidity, and high temperatures. As an example, human
ric. The Damerau metric is a generalization of the Levenshtein
cellular DNA undergoesanywherebetween 10-50breakagesin
distance which, in addition to deletions, insertions and substitu-
6 a cell cycle [23]. These DNA breakages or symbol/block dele-
tion errors also accounts for adjacent transposition edits.We first
1
provide constructions for codes that may correct either a single tionsresultinchangedstructuresofthestring:Ifastringbreaks
0
deletionorasingleadjacenttranspositionandthenproceedtoex- in two places, which is the most likely scenario, either the se-
2
tendtheseresultstocodesthatcansimultaneouslycorrectasingle quence reattaches itself without resulting in structural damage,
p deletion and multiple adjacent transpositions. We conclude with
reattaches itself in the opposite direction, resulting in what is
e constructionsfor joint block deletionand adjacent blocktranspo-
S sition error-correcting codes. called a reversal error, or the broken string degrades, resulting
inabursty(block)deletion;ifastringbreaksinthreepositions,
6
which is the second most likely breakage scenario, either the
I. INTRODUCTION
] adjacentbrokenblocksexchangepositionsoroneorbothblock
T The edit distance is a measure of similarity between two disintegrateleadingto a burstydeletion. Itis the latter scenario
I strings evaluated based on the minimum number of operations that motivates the study of channels in which adjacent blocks
.
s requiredto transformonestringintothe other.Iftheoperations of symbols may be exchanges or individual blocks deleted. It
c
[ are confined to symbol deletions, insertions and substitutions, is straightforward to see that this editing scenario corresponds
the distance of interest is the Levenshtein (edit) distance [15]. toa “blockversion”oftheDameraueditingprocess.Theblock
2
The Levenshtein distance has found numerous applications in editing processis hard to analyze directly, so we first study the
v
5 bioinformatics,where a weighted version of this metric is used symbol-level Damerau editing process and then proceed to an-
8 to assess the similarity of DNA strings and reconstruct phylo- alyze the block model. Also, for simplicity of exposition, we
8 genetic trees [14], and natural language processing, where the focus our attention on deletion and adjacent transposition er-
6 distanceisusedtomodelspellingerrorsandprovideautomated rors and delegate the more complex analysis of all four edit
0
word correction [4]. operations to future work.
.
1 Inparallelto theworkondevelopingefficientalgorithmsfor Our contributions are two-fold. We introduce the Damerau
0
computingtheeditdistanceandperformingalignmentsoflarge distance code design problem, and describe the first known
6
number of strings, a long line of results were reported on the scheme for correcting one deletion or one adjacent transpo-
1
: topic of designing codes for this distance function. Codes in sition. The scheme has near-optimal redundancy. We then
v
the edit distance are of particular importance for communica- proceed to extend and generalize this construction so as to
i
X tion in the presence of synchronization errors, a type of error obtain codes capable of correcting one deletion and one adja-
r encounteredinalmostallmodernstorageanddatatransmission cent transposition that also have near-optimal redundancy. Our
a systems.Classicalderivationsofupperboundsoncodesizesby results also shed light on the new problems of mismatched
Levenshtein [15] and single deletion-correcting code construc- Varshamov-Tenengoltz (VT) decoding and run length limited
tions by Varshamov and Tenengoltz [21], [22] have established VT codes. Second, we describe significantly more involved
the framework for studying many challenging problems in op- code constructions for correction of multiple adjacent trans-
timal code design for this metric [3], [7], [12], [18], [20]. position errors and proceed to introduce codes capable of
The Damerau distance is an extension of the Levenshtein correcting a block deletion and adjacent block transposition.
distance that also allows for edits of the form of adjacent sym- In the derivation process, we improve upon the best known
bol transpositions [4]. Despite the apparent interest in coding constructions for block deletion-correcting codes (i.e., codes
for edit channels, the problem of designing codes in the Dam- capable of correcting a block of consecutive deletions).
eraudistancewas notstudied before.A possiblereasonforthis The paper is organized as follows. Section II contains the
lack of interest in the Damerau distance may be attributed to problem statement and relevant notation. Section III contains
the fact that not many practical channel models involve adja- an analysis of the code design procedure for single deletion or
cent transposition errors, and even if they do so, they tend not single adjacenttransposition correction.Section IV contains an
to allowfor user-selectedmessages. Ourmotivatingapplication orderoptimalcodeconstructionfor correctinga single deletion
for studying codes in the Damerau distance is the emerging and a single adjacent transposition, as well a low-redundancy
paradigm of DNA-based storage [2], [6], [10], [25], [26]. In construction for codes correcting a single deletion and multi-
DNA-based storage systems, media degradation arises due to ple adjacent transpositions. Sections V and VI are devoted to
DNA aging caused by metabolic and hydrolitic processes, or our main findings: The best known code construction for sin-
gleblockdeletioncorrection,andcodescapableofcorrectinga Similarnotationwillbeusedforothertypesofeditingerrors,
single block deletion and a single adjacent block transposition. balls,distancesandcodes,withtheirmeaningapparentfromthe
context.Furthermore,fortheconvenienceofthereader,relevant
II. TERMINOLOGY AND NOTATION
notation and terminology referred to throughout the paper is
We start by defining the Damerau-Levenshtein distance,
summarized in Table I.
whicharoseintheworksofDamerau[8]andLevenshtein[15],
and by introducing codes in this metric. We then proceed to III. SINGLE TRANSPOSITION ORDELETION-CORRECTING
extend the underlying coding problem so that it applies to CODES
blocks, rather than individual symbol errors. Westartbydescribingageneralconstructionforsingletrans-
positionor deletion-correctingcodes.Wethenshowhowtouse
Definition1. TheDamerau–Levenshteindistanceisastring
this constructionin order to devise codes with near-optimalre-
metric, which for two strings of possibly different lengths
dundancy.
over some (finite) alphabet equals the minimum number of
insertions, deletions, substitutions and adjacent transposition
Construction 1 LetC (n,3)beasingle-error-correctingcode,
H
editsneededtotransformonestringintotheother.Theblock
and, as before, let C (n) be a single-deletion-correcting code.
D
Damerau–Levenshteindistancewithblocklengthbisastring
We define a code C (n), which we show in Lemma 2 is
T∨D
metric,whichfortwostringsofpossiblydifferentlengthsover
capable of correcting one transposition (T) or (∨) one deletion
some(finite)alphabetequalstheminimumnumberofinsertions,
(D) as follows:
deletions,substitutionsandadjacenttranspositioneditsofblocks
oflengthatmostbneededtotransformonestringintotheother. CT∨D(n)={x∈Fn2 :x∈CD(n),x∈CH(n,3)}.
For simplicity,we focuson editsinvolvingdeletionsand ad- Note that the code C (n) consists of codewordsthat belong
T∨D
jacent transpositions only, and with slight abuse of terminol- toasingledeletionerror-correctingcodeandhaveintegralsthat
ogy refer to the underlying sequence comparison function as belong to a single substitution error-correcting code.
the Damerau metric1. Furthermore, we restrict our attention to
binary alphabets only. Generalizations to larger alphabet sizes Lemma2.ThecodeC (n)fromConstruction1cancorrecta
T∨D
may potentially be accomplished by a careful use of Tenegoltz singleadjacenttranspositionorasingledeletion.
up-downencoding,describedin [1], [16], but thisproblemwill
Proof: We prove this claim by showing that for all x ∈
be discussed elsewhere.
For a vector x∈Fn, let B (x) denote the set of vectors CT∨D(n), onecan uniquelyrecoverx fromanyz ∈BT∨D(x).
2 T∨D Assume first that z ∈ Fn−1, so that z is the result of a
that may be obtainedfromx by either at most one single adja- 2
single deletion occurring in x. Since x ∈ C (n), one may
cent transposition (T) or at most one single deletion (D). Note D
apply the decoder of the code C (n) to successfully recover
that the size of B (x) is 2r(x), where r(x) is the number D
T∨D
x∈C (n).
of runs in x, i.e., the smallest number of nonoverlapping sub- T∨D
Assumethatz ∈Fn,sothatzistheresultofatmostonesin-
strings involving the same symbol that “covers” the sequence. 2
gle transposition in x. We show that d (x,z)61. When this
H
inequalityholds,sincexbelongstoacodewithminimumHam-
Example1. Suppose that x=(0,0,1,1,0)∈Fn. Then, mingdistance3,thevectorxcanbeuniquelydeterminedbased
2
on z. Note that since the mapping∂ is injective, d (x,z)=0
H
B (x)={(0,1,1,0),(0,0,1,0),(0,0,1,1),
T∨D if and only if x=z.
(0,0,1,1,0),(0,1,0,1,0),(0,0,1,0,1)}. Letthetransmittedwordxbesubjectedtooneadjacenttrans-
position involving the ith and (i+1)th bits, so that x 6=x
In particular, B (x) = B (x)∪B (x), where B (x) is i i+1
T∨D D T D and z = (x ,...,x ,x ,x ,x ,...,x ). First, we com-
the set of words obtained by deleting at most one element in 1 i−1 i+1 i i+2 n
pute the integral z as
x, while B (x) is the set of words obtained from at most one
T
adjacent transposition in x. n
z =(z ,z +z ,z +z +z ,..., z )=(z ,...,z ).
1 2 1 3 2 1 j 1 n
The derivative of x, denoted by ∂(x) = x′ is a vector de- j=1
X
fined as x′ = (x ,x +x ,x +x ,...,x +x ). Clearly,
1 2 1 3 2 n n−1 Let x = (x ,...,x ). Then, clearly (x ,...,x ) =
the mapping between x and x′ is a bijection. Hence, the in- 1 n 1 i−1
(z ,...,z ). Furthermore,
tegral ∂−1(x) , x is well-defined for all x ∈ Fn. For a set 1 i−1
2
X ⊆Fn,weuseX′ todenotethesetofderivativesofvectorsin i−1 i−1
2
X,andsimilarly,weuseX todenotethesetofintegralsofvec- zi = xj +xi+1 = xj +(1+xi)=1+x¯i,
tors in X. For two vectors x,y ∈Fn2, we let dH(x,y) denote Xj=1 Xj=1
theirHammingdistance.Furthermore,weletCH(n,d)standfor andforanyk >i+1,z = i−1x +x +x + k x =
any code of length n with minimum Hamming distance d, and k j=1 j i+1 i j=i+2 j
x , so that d (x,z)=1 as desired.
k H
similarly, we let C (n) stand for anysingle-deletion-correcting P P
D Notethatwe did notexplicitlystate in Construction1 which
code of length n.
codes to use. An obviouschoice would be a Hamming code as
thesinglesubstitutionerror-correctingcodeandtheVarshamov-
1Sinceweonlyconsiderdeletions, whatwerefertoasDameraudistance is
strictly speakingnotametric. Tenengoltz (VT) code as the single deletion error-correcting
Notation Description Positioninthemanuscript
BD(x) Thesetofwordsthatmaybeobtained fromatmostonesingledeletion inavectorx. EndofSection II.
BT(x) Thesetofwordsthatmaybeobtained fromatmostonesingleadjacent transposition inavectorx. EndofSection II.
BT∨D(x) BT∨D(x)=BD(x)∪BT(x). EndofSection II.
x′,∂(x) Thederivative ofx. EndofSection II.
x¯,∂−1(x) Theintegral ofx EndofSection II.
CH(n,d) AcodeofminimumHammingdistanced. EndofSection II.
CT∨D(n) Acodethatcancorrectasingleadjacent transposition ordeletion. Section III,Construction 1.
XD(n,a) Acodethatcancorrect asingledeletion error. SectionIII,preceding Claim1.
XH(n,a) Acodethatcancorrect asinglesubstitution error. SectionIII,preceding Claim1.
B(T,ℓ)(x) Thesetofwordsobtained fromxviaℓadjacent transpositions. Section IV,preceding Example2.
B(T,ℓ),D(x) Thesetofwordsobtained fromxviaℓadjacent transpositions andasingledeletion. Section IV,preceding Example2.
AVT-typecodetakenwithmodulusgivenbytheparameter ℓ.
CVT(n,a,ℓ) SectionIV,following Lemma6.
ThecodeCVT(n,a,b,ℓ)comprises asubsetofcodewords inCVT(n,a,ℓ)dictated bytheparameter b.
DVT,n,ℓ Adecoder forCVT(n,a,ℓ). SectionIV,following Lemma6.
DVT,n,b,ℓ Adecoder forCVT(n,a,b,ℓ). SectionIV,following Lemma8.
Acodewhichmaycorrectasingledeletion anduptoℓadjacent transpositions.
C(T,ℓ)∧D(n,a,b) SectionIV,beforeTheorem11.
C(T,ℓ)∧D(n,a,b)isasubsetofwordsinCVT(n,a,b,ℓ).
YT∧D(n,a1,a2) Acodeusedinthedefinition ofCT∧D(n,a1,a2). Section IV,following Corollary 12.
CT∧D(n,a1,a2) Acodethatmaycorrectoneadjacent transposition andonedeletion. Section IV,following Corollary 12.
BD,6b(x) Thesetofwordsthatmaybeobtained fromxviaaburstofconsecutive deletions oflengthatmostb. Section V-A,Part1.
BD,b(x) Thesetofwordsthatmaybeobtained fromxviaaburstofconsecutive deletions oflengthexactly b. Section V-A,Part1.
Cpar(n,b,d) Acodeusedtodetermine theweightofadeleted substring. Section V-A,Part1.
I(y,v,kI) Avectorobtained byinserting v intoy atpositionkI. Section V-A,preceding Claim 3.
D(y,b,kD) Avectorobtained bydeleting bconsecutive bits fromy startingatposition kD. Section V-A,preceding Claim 3.
Bal(n,b) A(balanced) setofwordsinwhichanysufficiently longsubstringhasroughlyhalfonesandhalfzeros. Section V-A,preceding Claim 4.
Acodefordetermining theapproximate location ofaburstofdeletions.
Codd(n,a,D) Section V-A,following Claim4.
b
ThecodeCVT(n,a,b,ℓ)comprises asubsetofwordsinCVT(n,a,ℓ).
SVTc,d(n,M) Acodefordetermining theexactlocation ofadeletion givenanapproximate location forthesame. Section V-A,Part3.
Acodewhichmaycorrect aburstofdeletions ofoddlength.
Codd(n,a,C,D) Section V-A,preceding Theorem17.
b ThecodeCbodd(n,a,C,D)isconstructed usingthecodes Cbodd(n,a,D)andSVTc,d(n,M).
Acodecapable ofcorrecting aburstofdeletions ofanylength6b.
Cb(n,a,C~,D~) Cb(n,a,C~,D~)isconstructed usingthecodeCbodd(n,a,C,D). Section V-B,following Example9.
BBT,b(x) Thesetofwordsobtained fromxviaoneadjacent blocktransposition. SectionVI,preceding Example11.
BBT∧D,b(x) Thesetofwordsobtained fromxviaoneadjacent blocktransposition andoneblockdeletion. Section VI,following Example12.
T(x,kT) Thevectorresulting fromtransposingthesymbolsatpositions kT andkT +1inx. Section VI,preceding Lemma22.
C(1) (n,a,C,D) Acodefordetermining theapproximate location ofablockofdeletions andadjacent transposition. SectionVI,following Lemma22.
TD,b
C(n,m;t1,t2) Acodeforcorrecting specialtypes ofbursterrors. SectionVI,following Definition 24.
COdd,B(n,a,C,D) Acodeforcorrecting anodd-length blockofdeletions andadjacent blocktransposition. SectionVI,following Lemma22.
b
CTD,b(n,a,C~,D~) Acodeforcorrecting oneblockofdeletions andoneadjacent blocktranposition. SectionVI,beforeTheorem27.
TABLEI
RELEVANTNOTATIONANDTERMINOLOGY.
code[15],oranycosetofthesecodes.Sincethecosetsofthese we use
codescoverFn,onecanseethatthereexistsacodewithredun-
2
dancy at most 2log(n+1). We show next how to improvethis
result by constructing one code that can serve both as a single
deletion-correctingfor x and a single error-correctingcode for n−1
x. The redundancy of this code is at most log n+log6. X (n,a)={x∈Fn : ix +
D 2 i
Our choice of codes is as follows. Let a be a non-negative Xi=1
integer such that 06a66n−4. For the single deletion code, (2n−1)xn ≡ a mod(6n−3)}.
For the code C (n,3), we choose Corollary5. There exists a single transposition or deletion-
H
correcting code whose redundancy is at most log(6n − 3)
n−2
X (n,a)={x∈Fn : (2i+1)x +(2(n−1)+1)x bits.
H 2 i n
Xi=1 Proof: Using the pigeon-holeprincipleconsidered in [21],
one may easily show that |X (n,a)| = |C (n,a)| > 2n ,
+(3(n−1)+1)xn−1 ≡amod(6n−3) . since C (a,n) partitions tHhe ambient spTa∨cDe Fn into 6n6n−−33
(cid:27) T∨D 2
codes, one of which has to have a size at least as large as the
right-hand side of the inequality.
Claim1. Foranyvectorx ∈ Fn,ifx′ ∈ X (n,a) thenx ∈ Note that every single transposition or deletion-correcting
2 D
X (n,a)andthusifx∈X (n,a)thenx∈X (n,a). code is also a single deletion error-correcting code. Hence, a
H D H
lowerboundonthe redundancyof thelatter codeis log n [13],
Proof: Suppose that x′ ∈XD(n,a). By definition, so that the difference between the redundancy of our dele-
tion/adjacent transposition codes and the redundancy of a
n−1
ix′ +(2n−1)x′ ≡amod6n−3. optimal single deletion code is at most log6 bits. We also note
i n
that improving the lower bound on a single transposition or
i=1
X deletion-correcting code is left as an open problem.
Therefore, since x′ = (x ,x +x ,x +x ,...,x +x ),
1 1 2 2 3 n−1 n
we have IV. CODES CORRECTING DELETIONSAND ADJACENT
TRANSPOSITIONS
n−1
x + i(x +x )+(2n−1)(x +x )≡a Wenowturnourattentiontothesignificantlymorechalleng-
1 i i−1 n−1 n
i=2 ing task of constructing codes that can correct both deletions
X
mod6n−3, and adjacent transpositions simultaneously. Our main result is
a construction of a code capable of correcting a single dele-
which implies that x∈XH(n,a). This proves the claim. tion along with multiple adjacent transpositions. At the end of
AccordingtoClaim1andLemma2,inordertoshowthatthe thissection,wepresentanimprovedconstructionforthespecial
codeCT∨D(n)=XD(n,a)isasingletranspositionordeletion- case of a single deletion and a single transposition.
correctingcode,we onlyhave to show thatthe codesXD(n,a) We start by introducing some useful notation. Let B(T,ℓ)(x)
and XH(n,a) have the desired error-correcting properties. denote the set of vectors that may be obtained by applying at
most ℓ adjacent transpositions (T) to x. Hence,
Lemma3. The codeX (n,a) is a single substitutionerror-
H
B (x)=B (...(B (x))...).
correctingcode. (T,ℓ) (T,1) (T,1)
ℓtimes
Proof: Let H = (3,5,7,..., 2n−3,3n−2,2n−1) so
that x ∈ X (n,a) if and only if HxT ≡ amod(6n −3). Let B(T,ℓ),D(x) denote t|he set o{fzvectors}that may be obtained
H fromxbyatmostℓadjacenttranspositionsfollowedbyatmost
Assume on the contrary that X (n,a) is not a single substitu-
H one single deletion. As before, let B (x) be the set of words
tion error-correctingcode.Then,there existtwo differentcode- D
that may be obtained by introducing at most one deletion into
words x ,x ∈ X (n,a) and two vectors e ,e such that
1 2 H j k x. With a slight abuse of notation, we use the same symbol B
x +e = x +e , where both e ,e have at most a single
1 j 1 k j k independentlyon the the argumentof the set beinga word or a
non-zero entry of value 1 or −1. Hence, we would have
collectionofwords.Inthelattercase,thesetBequalstheunion
H(x +e )T ≡H(x +e )T mod(6n−3), and of the corresponding sets of individual words in the argument.
1 j 2 k
The next example illustrates the relevant notation.
HeT 6≡HeT mod(6n−3),
j k
which holds if and only if e = e . Therefore, we must have Example2. Suppose that x=(0,0,1,1,0). Then,
j k
that x =x , a contradiction.
1 2 B (x)={(0,0,1,1,0),(0,1,0,1,0),(0,0,1,0,1)},
(T,1)
Lemma4.ThecodeXD(n,a)cancorrectasingledeletion. BD(x)={(0,0,1,1,0),(0,1,1,0),(0,0,1,0),(0,0,1,1)},
B (x)={(0,0,1,1,0),(0,1,1,0),(0,0,1,0),(0,0,1,1),
Proof: By definition, if x∈X (n,a), we may write (T,1),D
D
(1,0,1,0),(0,1,0,1),(0,1,0,0),(0,0,0,1)}.
HxT ≡amod6n−3,
where H = (1,2,3,...,n−1,2n−1). The result follows by Lemma6.Foranyx∈Fn2,
observingthat (1,2,3...,n−1,2n−1)is a Helbergsequence
B (x)=B (B (x))=B (B (x)).
(T,ℓ),D D (T,ℓ) (T,ℓ) D
asdefinedinDefinitionIII.2from[11].Thus,accordingtoThe-
orem III.4 of the same paper, the code X (n,a) can correct a Proof:Theproofisbyinductiononℓ.Forthebasecaseℓ=
D
single deletion. 1, we show that B (B (x)) = B (B (x)) by demon-
D (T,1) (T,1) D
The following corollary summarizes the main result of this strating that if y ∈ B (B (x)), then y ∈ B (B (x)).
(T,1) D D (T,1)
section. Furthermore if y ∈B (B (x)), then y ∈B (B (x)).
D (T,1) (T,1) D
Suppose that y(d) = (x ,...,x ,x ,...x ) is the re- x∈C (n,ℓ)fromx.Tothisend,definethefollowingcode:
1 i−1 i+1 n T∧D
sult of deleting the symbol at position i, where i ∈ [n]. Also, n
assume that y = y(d,t) is obtained from y(d) by transposing CVT(n,a,ℓ)={x∈bFn2 : ixi ≡amod(n+2ℓ+1)}.
the symbol in position j with the symbol in position j +1 in i=1
X
y(d), where j ∈ [n−2]. One needs to consider two different Since the code is a VT code, the decoder D for
VT,n,ℓ
scenarios: 1) j ∈[n−2]\(i−1); and 2) j =i−1. C (n,a,ℓ) can correct a single deletion occurring in any
VT
First, we show that if j ∈ [n − 2] \ (i − 1), then codewordinC (n,a,ℓ)[21].Notethatthestandarddefinition
VT
y ∈ BD(B(T,1)(x)). To see why this claim holds, note that if of a single deletion-correctingcode entails setting ni=1ixi to
j < i −1 then y may be generated by first transposing the be equal to some a modulo n+1 [21]. Our construction fixes
symbolsin positionsj,j+1 in x to obtainy(t) and thendelet- n ix to a modulo n+2ℓ+1 instead. As wePdemonstrate
ing the symbol in position i. Otherwise, if j >i, one may first iniC=l1aimi2, this change is needed due to the fact that adjacent
transpose the symbols in positionsj+1,j+2, and then delete tPranspositions may change the value of the syndrome a by at
the symbol in position i. Suppose now that j = i−1. Then most ±ℓ.
xi−i 6= xi+1 and so xi equals either xi−1 or xi+1. Suppose As before, and for the special case of VT codes, assume
that xi = xi−1. Then y may be generated by first transposing that x is the result of VT decoding the vector y where y ∈
xi and xi+1, and then deleting the symbol in position i−1. B(T,ℓ),D(x). Our first aim is to characterize the difference be-
Otherwise, if xi = xi+1, y may be obtained by first trans- tweenbx and x, and for this purpose we use an intermediary
posing xi−1 and xi and then deleting the symbol in position word y(ℓ) that is generated from at most ℓ adjacent transposi-
i+1. tionsinbx, i.e.,a wordsuchthaty ∈BD(y(ℓ)). Moreprecisely,
Using a similar argument, it can be shown that if we demonstrate that if both x,y(ℓ) ∈ C (n,a,ℓ), then the
VT
y ∈BD(B(T,1)(x)), then y ∈B(T,1)(BD(x)). This establishes decoder outputs DVT,n,ℓ(a,y) and DVT,n,ℓ(a,y(ℓ)) will differ
the base case BD(B(T,1)(x))=B(T,1)(BD(x)). only in the transpositions that actually occurred in x. On the
We now prove the inductive step. otherhand,ifx,y(ℓ) belongtotwodifferentVTcodes(i.e.they
Suppose that BD(B(T,ℓ)(x)) = B(T,ℓ)(BD(x)) holds for all have differentvaluesof the VT syndromeparametera), then x
ℓ < L. We show that BD(B(T,L)(x)) = B(T,L)(BD(x)) holds and x differ by at most 2ℓ adjacent transpositions. The follow-
as well. This may be seen from the following chain of equali- ing simple claim is a consequence of the fact that an adjacent
ties: transposition changes the VT syndrome by at most one.
b
BD(B(T,L)(x))=BD(B(T,L−1)(B(T,1)(x))) Claim2. Supposethat y(ℓ) = (y(ℓ),...,y(ℓ)) ∈ B (x)
1 n (T,ℓ)
=B(T,L−1)(BD(B(T,1)(x))) wherex∈Fn2.Then,onehas| ni=1ixi− ni=1iyi(ℓ)|6ℓ.
=B (B (B (x)))
(T,L−1) (T,1) D
Proof: The proof is by inPduction on ℓP. For the base case
=B (B (x)),
(T,L) D suppose y(1) ∈B (x). The result clearly holds if y(1) =x
(T,1)
and so assume y(1) is the result of transposing the symbols in
where the second line follows from the inductive hypothe-
positions j and j+1 in x. Then
sis, which is applied to each vector in the set, and where the
third line is a result of the previous result which showed that n n
BD(B(T,1)(x))=B(T,1)(BD(x)). | ixi− iyi(ℓ)|
Asaconsequenceofthepreviouslemma,wemayhenceforth Xi=1 Xi=1
assume that the deletionalways occursafter the adjacenttrans- = ix +(i+1)x − ix +(i+1)x
i i+1 i+1 i
position(s). We then say that a code C can correct ℓ adjacent
=|(cid:12)x −x |=1, (cid:16) (cid:17)(cid:12)
transpositions and a single deletion, and refer to it as a ℓ-TD (cid:12) i+1 i (cid:12)
(cid:12) (cid:12)
codeifforanytwodifferentcodewordsu,v ∈C,B (u)∩ since x 6=x . For the inductive step, suppose that the result
(T,ℓ),D i i+1
B (v) = ∅. Our code construction and the ideas behind holds for all ℓ < L and consider the case ℓ = L. Let y(L) ∈
(T,ℓ),D
the coding approach are best explained through the decoding B andlety(L−1) ∈B besuchthaty(L) andy(L−1)
(T,L) (T,L−1)
procedure. differ by at most one single adjacent transposition. Then,
Suppose that the code CT∧D(n,ℓ) is an ℓ-TD code, which n n
is a subset of codewords of a single deletion-correcting code. | ix − iy(L)|
i i
Assume also that x ∈ CT∧D(n,ℓ) was transmitted and that Xi=1 Xi=1
the vector y was received, where y is the result of at most n n n n
ℓ transpositions followed by at most one single deletion in x. =| ixi− iyi(L−1)+ iyi(L−1)− iyi(L)|
The simplest idea to pursue is to try to correct the single dele- i=1 i=1 i=1 i=1
X X X X
n n n n
tion by naively applyingthe decoderfor the chosen constituent 6| ix − iy(L−1)|+| iy(L−1)− iy(L)|
single-deletioncode.Clearly,suchadecodermayproduceaner- i i i i
i=1 i=1 i=1 i=1
roneousresult due to the presence of the adjacenttransposition X X X X
6L−1+1=L.
errors.ItisthereforeimportanttoconstructthecodeC (n,ℓ)
T∧D
in such a way that the result of the “mismatched” deletion cor-
rectionx, obtainedfromy, iseasyto characterizeandcontains As a consequenceof the previousclaim, if x∈C (n,a,ℓ)
VT
onlyalimitednumberoferrorsthatmaybecorrectedtorecover and y(ℓ) ∈ B (x), then y(ℓ) ∈ C (n,aˆ,ℓ) for some aˆ,
(T,ℓ) VT
b
where |a−aˆ| 6 ℓ. The next lemma summarizes the previous (n+2ℓ+1).ThedecoderD forC (n,a,ℓ)producesthe
VT,n,ℓ VT
discussion. vector x ∈ C (n,a,ℓ) by inserting a zero into the first posi-
VT
tionk′ thathasa′ onestotherightofit.Ifx =0,thenclearly
k
Lemma7. Suppose that y(ℓ) ∈ B(T,ℓ)(x), where x ∈ a′ = vRb, and the decoder correctly outputs x so that x = x.
C (n,a,ℓ),andlety ∈B (y(ℓ)).Then,D (aˆ,y)=y(ℓ) If the decoder D for C (n,a+v,ℓ) were applied to y
VT D VT,n,ℓ VT,n,ℓ VT
forsomeaˆsuchthat|a−aˆ|6ℓ. instead, one would have b
n−1
Example3. Suppose that x = (0,1,1,0,0,1,0,0,0,0,1,0) ∈
a′′ ≡a+v− iy mod(n+2ℓ+1)≡a′+v mod(n+2ℓ+1).
i
C (12,3,3)wastransmittedandthatthevectory =(0,1,1,0,
VT
i=1
0,1,0,0,1,0,0)wasreceivedafteratmostthreeadjacenttrans- X
positions and a single deletion. For y(3) =(0,1,1,0, 0,1,0,0, Hence, the decoder DVT,n,ℓ for CVT(n,a+v,ℓ) would insert
0,1,0,0) (where y ∈ B (y(3)), we have n−1y ≡ 2mod a zero in the vector y at the first position k′′ that has a′ +v
D i=1 i
19.Thus,sincea=3andaˆ=2,wegetthat|a−aˆ|616ℓ=3 ones to the right of it. The claim follows by observing that the
as desired. P position that has a′+v ones after it is in the same run as the
Note that if we use the decoder DVT,12,3 we arrive position in y with (vL−v) ones preceding it.
at x = D (3,y) = (0,1,1,0,0,0,1,0,0,1,0,0). Suppose next that y is the result of deleting a one from
VT,12,3
Hence, we have x = (0,1,1,0,0,0,1,0,0,1,0,0), and x ∈ C (n,a,ℓ). Let a′ ≡ a− n−1iy mod(n+2ℓ+1).
VT i=1 i
x=b(0,1,1,0,0,1,0,0,0,0,1,0). The decoder DVT,n,ℓ for CVT(n,a,ℓ) producesthe vector x∈
C (n,a,ℓ)byinsertingaoneinPtothefirstpositionk′ thathas
b VT
We characterize next the difference between DVT,n,ℓ(a,y) a′−k′ ones to the right of it. If xk = 1, then clearly k′ =b k
and D (aˆ,y) for the case that |a−aˆ| 6 ℓ, as the value
VT,n,ℓ and the decoder correctly outputs x, so that x=x.
aˆ is not known beforehand.Our main result may be intuitively
If the decoder D for C (n,a+v,ℓ) were applied to
VT,n,ℓ VT
described as follows: Suppose that y ∈ B (x), where x ∈
D y instead, then one would have a′′ ≡a′+v mbod(n+2ℓ+1)
C (n,a,ℓ) and where y is obtained by deleting the kth bit,
VT as before. The decoder D for C (n,a+v,ℓ) would in-
VT,n,ℓ VT
x , from x. Also, assume that the value of x is known to the
k k sert a one into the vector y at the first position k′′ that has
decoder and that x = D (a+v,y), for some offset v, is
VT,n,ℓ a′ + v − k′′ ones to the right of it. This produces a vector
obtained by inserting the bit x into y at some position de-
k x. Since the first position k′′ in y with a′ + v − k′′ ones
termined by the dbecoder. Then, if xk = 0, we may obtain x to the right is the same as the first position in y following
from x by sliding the inserted bit to the left/right using a se-
(bv + k − vL − 1) zeros, the claimed result also follows for
ries of adjacent transposition operations past at most v ones.
x =1. (cid:4)
k
Otherwbise, if xk =1, then we can obtain x from x by sliding
the inserted bit to the left/right past at most v zeros. The next
The previous lemma motivates the introduction of a modifi-
lemma rigorously summarizes this observation.
b cationofVTcodes,whichwillbeusedasaconstituentcompo-
nentin a constructionof codescapable of correctinga deletion
Lemma8.Supposethatyistheresultofasingledeletionoccur-
and multiple adjacent transpositions. This modified code struc-
ringinx ∈ C (n,a,ℓ)atpositionk.Givenk,letv = |{j ∈
VT L ture also leads to a straightforward decoding procedure of the
[n] : j < k,x = 1}|andv = |{j ∈ [n] : j > k,x = 1}|.
j R j underlying codes. The code may be defined as follows:
Then,
1) Ifxk = 0,thenforallv ∈ {−vR,−vR +1,...,vL},one CVT(n,a,b,ℓ)={x∈Fn2 : (1)
mayobtainDVT,n,ℓ(a+v,y)byinsertingthesymbol0into n
yimmediatelyafterthe(v −v)-thone. ix ≡amod(n+2ℓ+1),
L i
2) Ifxk = 1,thenforallv ∈ {−(k−1)+vL,−k+vL+2, Xi=1
...,(n−k)−v },onemayobtainD (a+v,y)by n
R VT,n,ℓ
insertingthesymbol1intoyimmediatelyafterthe(v+k− xi ≡bmod2}.
vL−1)-thzero. Xi=1
The decoder for C (n,a,b,ℓ), denotedby D , oper-
VT VT,n,b,ℓ
Example4. Suppose that x = (0,1,1,0,0,1,0,0,0,0,1,0) ∈
ates as follows. Suppose that x∈C (n,a,b,ℓ) is transmitted
VT
C (12,3,3),andthatxˆ =D (3,y),wasobtainedbyVT
VT VT,n,ℓ and that y ∈B (x) is received. Suppose that n denotes
(T,ℓ),D 1
decoding y = (0,1,1,0,1,0,0,0,0,1,0). For v = 2, one has the number of ones in y. Then, for a ∈ Z and b ∈ F ,
n+2ℓ+1 2
D (5,y)=(0,0,1,1,0,1,0,0,0,0,1,0),whereasforv =
VT,n,ℓ D (a,y) executes the following steps:
VT,n,b,ℓ
−1, one has D (2,y)=(0,1,1,0,1,0,0,0,0,0,1,0).
VT,n,ℓ 1) Set x≡ n−1y +bmod2.
Next, suppose that y = (0,1,1,0,0,0,0,0,0,1,0) – y is i=1 i
2) Compute a′ ≡a− n−1iy mod(n+2ℓ+1).
the result of deleting the third 1 at position k = 6 from x = P i=1 i
3) Ifx=0anda′ ∈{0,1,...,n },insertazerointothefirst
(0,1,1,0,0,1,0,0,0,0,1,0).Inthiscase,choosingv =3gives 1
position in y that hPas a′ ones on its right.
D (6,y) = (0,1,1,0,0,0,0,0,1,0,1,0), while v = −2
VT,n,ℓ
If a′ ∈ {n +1,n +2,...,n +ℓ}, insert a zero in the
gives D (1,y)=(0,1,1,1,0,0,0,0,0,0,1,0). 1 1 1
VT,n,ℓ
first position in y.
ProofofLemma8: Supposefirstthaty isthe resultofdelet- If a′ ∈{n+ℓ+1,n+ℓ+2,...,n+2ℓ}, insert a zero in
ing a zero fromx∈C (n,a,ℓ). Leta′ ≡a− n−1iy mod the last position of y.
VT i=1 i
P
4) If x=1 and a′ ∈{n +1,n +2,...,n}, insert a one in BasedontheresultsonmismatchedVTdecodingandCorol-
1 1
the first position k of y that has a′−k ones to its right. lary10,wearenowreadytodefineafamilyofcodescapableof
Otherwise, if a′ ∈{n+1,n+2,...,n+ℓ}, insert a one correcting a single deletion and multiple adjacent transposition
in the last position of y. errors.Recallthatgivena binarywordx, its derivative∂(x)=
If a′ ∈ {n −ℓ+1,n −ℓ+2,...,n }, insert a one in x′ is defined as x′ = (x ,x +x ,x +x ,...,x +x )
1 1 1 1 2 1 3 2 n n−1
the first position of y. and its inverse as ∂−1(x) = x = (x ,x +x ,..., n x ).
1 1 2 i=1 i
Note that the VT decoder discussed so far aims to correct We claim that the code C(T,ℓ)∧D ⊆Fn2, defined as P
a single deletion only, but potentially in a mismatched fash- C (n,a,b)={x∈Fn : x∈C (n,4ℓ+1),
(T,ℓ)∧D 2 H
ion as additional adjacent transposition errors may have been
x∈C (n,a,b,ℓ)} (2)
incurredduring deletion correction. The outputof the deletion- VT
correctingdecoderhastobefedintotheinputofatransposition is an ℓ-TD code (i.e., a code capable of correcting ℓ adjacent
error-correctingcode, and we describe how this subsequentde- transpositions (T,ℓ) and (∧) one deletion (D)).
codingisaccomplishedafterprovidinganillustrationoftheVT
decoding process. Theorem11.ThecodeC (n,a,b)isanℓ-TDcode.
(T,ℓ)∧D
Proof: Suppose that y ∈ B (x). We show how to
Example5. Suppose that x = (0,1,1,0,0,1,0,0,0,0,1,0) ∈ (T,ℓ),D
recover x from y. First, we determine x = D (a,y).
C (12,3,0,3), and that y = (0,1,1,0,1,0,0,0,1,0,0) VT,n,b,ℓ
VT
From Corollary 10, we have that x ∈ B (x). Since x ∈
is the received word, which is the result of a single dele- (T,2ℓ)
tion and a single transposition. We first apply the decoder B(T,2ℓ)(x), we have dH(∂−1(x),x) 6 2bℓ. Because the mini-
D to y. In the first step of the procedure, we mum distance of the code C(T,ℓ)∧D(n,a,b) is 4bℓ+1, we can
coVnTcl,u12d,e0,3that the deleted bit has value x = 0. In the uniquelybrecover x from ∂−1(xb).
second step of decoding, we compute a′ = 3. Since
Corollary12. Thereexistsanℓ-TDcodewhichredundancyat
0 6 a′ 6 4, we have x = (0,1,0,1,0,1,0,0,0,1,0,0). b
most2ℓlog n+log(n+2ℓ+1)bits.
Note that x = (0,1,0,1,0,1,0,0,0,1,0,0), and x =
(0,1,1,0,0,1,0,0,0,0,1,0b), differ in two adjacent transposi- Next,we improveuponthisresultforthe case whenℓ=1.Let
tions. b a ,a ∈Z . Define Y (n,a ,a )⊆Fn according to
1 2 n+2L+1 T∧D 1 2 2
The previous example illustrates that x and x differ in a Y (n,a ,a )={x :x =0,
T∧D 1 2 n
limited numberoftranspositionswhichdependsonthe original n−1
number of transposition errors. In particular, for bthe given ex- (2i+1)xi ≡a1 mod(n+2L+1),
ample, the two vectors differ in two adjacent transpositions as i=1
X
x is the result of a single deletion and a single transposition in n−1
y. The nextlemma givesa moreprecisecharacterizationof the (2i+1)2xi ≡a2 mod(n+2L+1)},
“bdistance” between x and x. Xi=1
where L > 1 is chosen so that n+2L+1 is a prime number
Lemma9. Suppose that by(ℓ) ∈ B(T,ℓ)(x) where x ∈ greater than 2n−1. Let
CVT(n,a,b,ℓ)andwherey ∈BD(y(ℓ)).Letx=DVT,n,b,ℓ(a,y). C (n,a ,a )=Y′ (n,a ,a ),
Thenthefollowingstatementsaretrue: T∧D 1 2 T∧D 1 2
whereY′ standsforthecollectionofallderivativesofwordsin
1) Ifxistheresultofinsertingazeroinybinapositionwith
v onestotherightoftheinsertedbit,theny(ℓ)canbeob- Y. We have the following lemma.
R
tainedfromy byinsertingazeroiny inthefirstposition
b Lemma13.Foralla ,a ∈Z ,thecodeC (n,a ,a )
withj onestotherightofitwherej ∈ {v −ℓ,v −ℓ+ 1 2 n+2L+1 T∧D 1 2
R R
isa1-TDcode.
1,...,v +ℓ}.
R
2) Ifxistheresultofinsertingaoneiny inpositionk with
Proof: We use the same approach as the one outlined in
v onestotherightoftheinsertedbit,theny(ℓ)canbeob-
R the proof of Claim 1.
tainbedfromybyinsertingaoneinyinthefirstpositionj Since n−1(2i+1)x ≡a mod(n+2L+1) and x =0,
withj onestotherightofitwherej +j ∈ {k+v − i=1 i 1 n
1 1 R we have that
ℓ,k+v −ℓ+1,...,k+v +ℓ}. P
R R n
ix′ ≡a modn+2L+1. (3)
The following corollary summarizes one of the main results of i 1
i=1
this section. X
Furthermore, since x =0,
n
n
Corollary10. Suppose that y ∈ B (x) where x ∈
(T,ℓ),D x′ ≡0mod2. (4)
C (n,a,b,ℓ) and let x = D (a,y). Then x ∈ i
VT VT,n,b,ℓ
i=1
B (x). X
(T,2ℓ)
From (3) and (4), it is clear that if x∈ Y (n,a ,a ), then
b T∧D 1 2
Consequently, the mismatched VT decoder increases the num- x′ ∈ C (n,a ,0,L). Similarly to what was done in Theo-
b VT 1
ber of adjacent transposition errors by at most a factor of two. rem11, itcanbeshownthatifL>1 andY (n,a ,a )has
T∧D 1 2
Hamming distance at least 5, then C (n,a ,a ) is a 1-TD is fixed. It will be helpful to think of codewords of length
T∧D 1 2
code.Bydesign,L>1andsoweturnourattentiontoshowing n = cb, c > 1, as two dimensional arrays formed by writing
that Y (n,a ,a ) has Hamming distance at least 5. the bits in the codeword column-wise, i.e., by placing the bits
T∧D 1 2
We claim that the vectors in Y (n,a ,a ) represent a (x ,x ,...,x ) in an orderly fashion within the first column
T∧D 1 2 1 2 b
coset of a Berlekamp code [17, Chapter 10.6] with Lee dis- of the array, the bits (x ,x ,...,x ) within the second
b+1 b+2 2b
tance 5, which implies the desired result. To prove the claim, column and so on. As an example, for c =n/b, the codeword
notethatthebinarycodeY (n,0,0)hasa parity-checkma- x=(x ,...,x ) would read as follows:
T∧D 1 n
trix of the form
x x x ... x
1 b+1 2b+1 c(b−1)+1
3 5 7 ... 2n−1
H = 32 52 72 ... (2n−1)2 . x2 xb+2 x2b+2 ... xc(b−1)+2. (5)
.................................
(cid:20) (cid:21)
According to [17, Chapter 10.6], in order for YT∧D(n,0,0) to xb x2b x3b ... xn 
have minimum Lee distance 5, the following statement has to  
Forsimplicity,throughoutthe remainderofthissection,weuse
be true: For any two columns of H, say h and h , it has to
i j the term “interleaved sequence” to refer to a row in the array.
hold that
Note that in this setting, a block of b consecutive deletions in
0
hi1 +hi2 6= c , a codeword x leads to one deletion within each interleaved se-
(cid:20) (cid:21) quence, and that the locations of deletions in the interleaved
for anypossible choiceof c∈Fn+2L+1. Clearly, this condition sequences are correlated. As an example, the block may cause
istruesincen+2L+1isanoddprimeandthesumoftwoodd the same deletion location in the first interleaved sequence, but
numberscannotequalanotheroddnumber.Thus,YT∧D(n,0,0) affectthesymbolsintheotherinterleavedsequencesdifferently
has minimum Lee distance at least 5 and so YT∧D(n,a1,a2) (The deleted symbols are underlined):
has minimum Lee distance at least 5, as claimed.
x x x ... x
Theaboveconstructionimprovesuponthegeneralconstruction 1 b+1 2b+1 c(b−1)+1
x x x ... x
describedbytheresult(2)intermsoflog nbitsofredundancy.  2 b+2 2b+2 c(b−1)+2, (6)
.................................
Remark1. It has been a long standing open problem to find xb x2b x3b ... xn 
extensions for the single-deletion VT code construction which or  
would have order optimal redundancy and impose syndrome
x x x ... x
1 b+1 2b+1 c(b−1)+1
constraintsoftheform f (i)x ≡a mod (n+1),forsome
judiciouslychosenfunctioinskf (i)i.Attemptsbasedonusingthis x2 xb+2 x2b+2 ... xc(b−1)+2, (7)
k .................................
P
approach have failed so far [3]. On the other hand, the result
of Lemma 13 shows that syndrome constraints of the form de- xb x2b x3b ... xn 
or  
scribed above can accommodate combinations of one deletion
and other forms of errors, such as adjacent transpositions. x x x ... x
1 b+1 2b+1 c(b−1)+1
x x x ... x
 2 b+2 2b+2 c(b−1)+2. (8)
Corollary14. Thereexistsa1-TDcodewhichredundancyat .................................
most2log n+cbits,forsomeabsoluteconstantc. xb x2b x3b ... xn 
 
In the next section, we turn our attention to the problem As a result, by finding the location of the deletion in the first
of constructing codes capable of correcting transposition and interleaved sequence does not automatically allow one to de-
deletion errors in the form of blocks of bits. First, we analyze termine the “shift” of the block with respect to that location.
the problem of constructing codes capable of correcting a sin- Furthermore, deletion correcting codes such as VT codes only
gle blockof adjacentdeletions.Then,we focusonconstructing identify the run of symbols in which the deletion occurs and
codes capable of correcting a single transposition of adjacent not its exact position, as the goal is to reconstruct the correct
blocks in addition to handling one block deletion. codewordand not precisely determine the location of the error.
As a result, furtheruncertaintyexists aboutthe locationsof the
V. CODES FORCORRECTING A BLOCK OFDELETIONS deletions in the second, third etc. interleaved sequence of the
We describe next a new family of codes capable of correct- codeword.
ing one block of at most b consecutive deletions; the codes re- To mitigate these problems, the authors of [5] proposed a
quirelogb logn+O(b2 log b loglogn)bitsofredundancy,and construction of codes capable of correcting a block of consec-
hence improve upon the state-of-the art scheme which requires utive deletions of length exactly b based on imposing simple
at least (b − 1)logn bits of redundancy [19]. The proposed constraintson the interleavedsequencesof a codeword.A con-
block-deletion codes will subsequently be used in Section VI structionwithredundancyofapproximatelyblognbitsrequires
to construct codes capable of correcting both a block of dele- all the interleaved sequences of (18) to belong to a VT code.
tions (which we alternatively refer to a burst of deletions) and Themaindrawbackofthisconstructionisthateachinterleaved
anadjacenttranspositionoftwoblocksofconsecutivesymbols. sequence is treated independently of the others and that con-
To explain the intuition behind our approach, we start with sequently, the redundancy of the codes is too high. To address
a short overview of existing code constructions for correcting this problem, one should use the position of the deletion in
a block of consecutive deletions, where the length of the block the first interleaved sequence to approximately determine the
location of the deletion in the second row and similarly for A. Odd Length Blocks
all other subsequent rows. In [5], the authors also proposed a
Our code construction is centered around three main ideas:
code which has an alternating sequence (i.e., a sequence of the
1) The use of VT codes (10).
form 0,1,0,1,0,1,...) as its first interleaved sequence and all
2) The use of running sum constraint (12).
the remaining interleaved sequencessatisfying a constraint that
3) The use of a sequence of Shifted VT codes [19], defined
requires log3 bits of redundancy. The proposed code may be
in (15) i.e., codes that enforce multiple modular VT-type
easily decoded by first determining the location of the deletion
constraints with parameter values smaller than n+1.
in the first row through a reference to the alternating sequence
structure. Then, this location is used by the remaining rows to As discussed in more details in what follows, our choice of
correct the remaining b−1 deletions. This approach requires the Shifted VT codes requires approximately b2loglogn bits
at least n/b bits of redundancy, due to the fact that one has to of redundancyand the constrained coding constraint requires a
fixthefirstrowofthecodewordarray.Thus,theredundancyof single bit of redundancy, the proposed construction introduces
this approach is actually higher than that of the code that uses roughly logn+b2loglogn bits of redundancy.
individual VT code constraints for each interleaved sequence. Thedecoderoperatesasfollows.Supposethaty istheresult
of deleting t consecutive bits from x, with t 6 b and t odd.
The alternating sequence approach was improved and gen- Then,
eralized in [19], where the authors constructed block deletion-
1) Thedecodercomputesanumberofparitiesanddecideson
correcting codes with a significantly more relaxed constraint
theappropriateShiftedVTcode(15)touseindetermining
placedonthefirstinterleavedsequence.Theirideawastocom-
the Hamming weight of the bits deleted from x.
bine constrained coding with a variant of VT codes which we
2) Using both the VT-type constraint (10) and the constraint
explainindetailsinwhatfollows.Therelaxedconstraintsallow
(12), the decoder determines an approximate location for
one to approximately determine the locations of the remaining
the block deletion in x that resulted in y.
deletions in x after decoding the first interleaved sequence of
3) Given the approximate location of the block of deletions,
the array. The constrained and VT-type constraints imposed on
the decoder uses a series of Shifted VT codes (15) to de-
the higher index rows nevertheless allow for unique recovery
termine the exact locations and values of the bits deleted
of the codewordx by using VT codes confinedto the “suspect
from x that lead to y.
range”predictedto harborthe deletions.The codesconstructed
Part 1. Determining the weight of the deleted substring.
in [19] require approximately logn bits of redundancy for the
We start with some relevant terminology and notation. For a
constraint in the first row of the array, and loglogn bits of
word x ∈ Fn, let B (x) denote the set of all words that
redundancy for each of the remaining rows. This results in a 2 D,6b
maybe obtainedfromx by deletingatmostb consecutivebits.
total redundancy of roughly logn+(b−1)loglogn bits for
For example, for x=(0,1,1,0,0,1)∈F6, we have
correcting a block of consecutive deletions of length exactly b 2
(compared to the redundancy of [5] which equals blogn bits).
B (x)= (0,1,1,0,0,1),(1,1,0,0,1),(0,1,0,0,1),
D,62
To allow for correctingany single block of length at most b, n(0,1,1,0,1),(0,1,1,0,0),(1,0,0,1),(0,0,0,1),
the codes from[19] have to be changedso as to includenested
(0,1,0,1),(0,1,1,1),(0,1,1,0) .
redundantbitsthatcapturemultiplecodingconstraintsandmay
allow for correcting a range of block lengths. Which of the o
Similarly, let B (x) denote the set of words that may be ob-
constraints to use is apparent upon observing the length of the D,b
tained from x by deleting exactly b consecutive bits.
received word: To correct one block of any possible length at
Furthermore, given a vector d ∈ Fb, define the code
mostb,thedecoderfortheunderlyingcodelocatestheposition 2
C (n,b,d) as follows2:
of the blockof consecutivedeletionsdifferentlyfor each possi- par
bleblocklength.Forinstance,if xexperiencesa blockerrorof
length b1 6 b, then the code uses one VT-type constraint, say ⌈n−bj+1⌉
KVT,b1. However, if x experiences an error burst of length b2 Cpar(n,b,d)={x∈Fn2 :∀j ∈[b], xj+bi ≡dj mod2}.
with b2 <b1, then thecodeeffectivelyusesa differentVT-type i=1
X
constraint, say K . Note that since each of the constraints
VT,b2 Itisstraightforwardtoseethatthecodeimposesasingleparity-
K , 2 6 i 6 b, is de facto a VT-type constraint, one re-
VT,bi check constraint on the interleaved sequences of x, which suf-
quires roughly (b−1)logn+b2loglogn bits of redundancy,
fices to determine the weight of the deleted block. In addition,
compared to the b2log n redundancy which would have been
we observe that we used a set of parameters d for the weight
required by the scheme in [5]. i
constraints, rather than the classical even parity constraints for
reasonsthatwillbecomeapparentinthesubsequentexposition.
Our approach in this work for a further improvement is to
In a nutshell, the resulting codes of the section will be nonlin-
reuse the same VT-type constraint for multiple possible block
ear and averaging arguments for the size of codes require the
lengths, in which case the redundancy will amount to roughly
use of a range of parameter values.
logb logn+logbb2loglogn bits. To describe this method,we
start with a construction that allows for correcting one odd-
2We use the subscript par to refer to the function of the code, which is
length block of consecutive deletions of length at most b, and
to recover the weight of the deleted block (substring) by using simple parity
then proceed to extend the result for even-length blocks. checks.
Example6. Suppose that x = (0,1,1,0,0,1,0,1,0,1,0,1) ∈ toB/2andif bisodd,thenonecanshow(seeLemma15)that
C (12,2,(1,1))wastransmittedandy =(0,1,1,0,0,1,0,1,0,1f)orany i ,i such that i −i >B, x,u do notboth belongto
par 1 2 2 1
∈B (x) was received instead. Since x ∈C (12,2,(1,1)), thesameVTcode.Therefore,whenablockofdeletionsoccurs,
D,2 par
it is straightforwardto determinethat the bits 0,1 were deleted Lemma 15 shows that it will be possible to determine approxi-
from x to obtain y. Notice, however, that we cannot infer mately (to within B bits) the location of the block of deletions
the order in which the deleted bits {0,1} appeared in x from by attempting to insert a block of bits into different positions
the constraints of the code C (12,2,(1,1)), nor their exact andcheckwhethertheyleadto avectorthatsatisfies aVT-type
par
location. constraint. According to Lemma 15, if the VT-type constraint
is satisfied, we know the location of the block of deletions to
Part 2. Imposing the generalized VT conditions.
within B positionsand we can determineexactlythe value and
Given y ∈ Fn−b, v ∈ Fb and k ∈ [n − b + 1], let
2 2 I location of the deleted bits using the Shifted VT codes in (15).
I(y,v,k )∈Fn bethevectorobtainedbyinsertingv intoy at
I 2 For notational convenience, we henceforth assume that n is
position k . For instance, if y = (0,1,1,0,0,1)∈ F6, k = 1
I 2 I a poweroftwo so thatlogn is a positiveinteger.LetBal(n,b)
and v = (0,1), then I(y,v,1) = (0,1,0,1,1,0,0,1) ∈ F8.
2 denote the following “balanced” set of sequences:
Similarly, let D(y,b,k ) be the result of deleting b consec-
D
utive bits from y starting at position kD. Thus, D(y,2,2) = Bal(n,b)= x∈Fn :∀B ∈[n],B >b4logn,∀j ∈[n−B+1],
(0,0,0,1) ∈ F4. As before, let wt(x) stand for the Hamming 2
weight of a vec2tor x. nB B j+B−1 B B
− < x < + . (12)
i
2 3b 2 3b
Claim3.Letv1,v2 ∈ Fb2,andsupposethaty ∈ F2n−bandthat Xi=j o
w1 = wt(v1) = wt(v2),w2 = wt(yi1,...,yi2−1),wherei1 < We have the following claim, the proof of which may be
i2.Letx=I(y,v1,i1)∈Fn2 andu=I(y,v2,i2)∈Fn2.Then, found in Appendix A.
n n
iu − ix =(i −i )w −bw +δ, (9)
i i 2 1 1 2 Claim4.Forapositiveintegern>10andb>5,
i=1 i=1
X X
where|δ|<b2. log|Bal(n,b)|>n+log 1−2n2−92b2loge .
(cid:16) (cid:17)
Remark2. Note that the term δ arises due to the fact that the Note that as a consequence of Claim 4, we have
sequences v and v have the same weight but potentially dif-
1 2
ferent locations of their nonzero symbols. log|Bal(n,b)|>n−1, (13)
Let C (n,a,b) be a VT code of the form and the coding constraint incurs not more than one bit of re-
VT,b
dundancy.
n
CVT,b(n,a,b)={x∈Fn2 : ixi ≡amod(bn+b2)}. (dLe,t..D.,d=)∈(dF1i,.dD2,e.fi.n.e,db), where for i ∈ [b], di =
i=1 1,i i,i 2
X
(10)
n
Clearly,two vectorsx andu ofthe formas definedin Claim 3 Codd(n,a,D)={x∈Fn : ix ≡amodbn+b2,
b 2 i
cannot both lie in the same code capable of correcting a block i=1
X
of deletions of length at most b. To see this, note that y = x∈Bal(n,b),∀i∈[b]
D(x,b,i1)=D(u,b,i2)∈BD,b(x)∩BD,b(u),acontradiction. x∈Cpar(n,i,di)}.
Ifweassumethatx,u∈C (n,a,b)aretypicalsequences
VT,b
generated by an i.i.d uniform source, then, with high probabil- We show next that if y is the result of an odd-length block
ity, w will be close in value to i2−i1. In order for x and u of deletions occurring in x∈Codd(n,a,D) starting at position
2 2 b
to belong to different codes (i.e., codes with different VT syn- k ,thenthereexistsadecoderforCodd(n,a,D)thatiscapable
D b
dromes)capableofcorrectingblockdeletionsoflengthatmost ofproducinganestimate,saykˆ ,forthestartingpositionofthe
D
b for an overwhelming large portion of the constituent vectors blockof deletions,with |k −kˆ |<b4logn. We thenproceed
D D
v, based on Claim 3 and the definition of CVT,b(n,a,b), we to explain how to recover the exact value and location of the
need to ensure that the right-hand side of (9) is not zero, i.e., deleted bits using the SVT codes of (15).
that First, we show that if y ∈ B (x) and x ∈ Codd(n,a,D),
D,t b
i −i with t 6 b an odd integer, one can obtain a good estimate for
(i −i )w +δ 6=b 2 1. (11)
2 1 1 2 the location of the block of deletions given that we know the
Hamming weight of the bits that were deleted (we can obtain
Note that if b is odd, then w cannot be equal to b/2.
1
thisfromadecoderforthesubcodeC (n,i,d )). Thefollow-
Next, we construct a codebook that ensures that (11) is sat- par i
ing result, similar to Claim 3, describes the relevant properties
isfied for any choice of distinct (code)words. The idea is to
of the decoder.
construct a set of codewords x with the following property:
Every block (substring) in x of length i − i = B, where
2 1
B >b4logn,hastohaveHammingweightapproximatelyequal Lemma15.Letv ,v ∈ Ft,wt(v ) = wt(v ),y ∈Fn−t,and
1 2 2 1 2 2
to B. If everyblockof B >b4logn bitsin x has weightclose supposethattisanoddnumberwheret6bandb>5.Fori <
2 1